Question

Find the probability distribution of
(i) number of heads in two tosses of a coin.

Answer 1

(i) When one coin is tosses twice, the sample space is $\{HH,HT,TH,TT\}$
Let $X$ represent the number of heads.
$\therefore X\left(HH\right)=2,X\left(HT\right)=1,X\left(TH\right)=1,X\left(TT\right)=0$
Therefore, $X$ can take the value of $0,1$ or $2$.
It is known that,
$P\left(HH\right)-P\left(HT\right)-P\left(TH\right)-P\left(TT\right)=\frac{1}{4}$
$P(X=0)=P\left(TT\right)=\frac{1}{4}$
$P(X=1)=P\left(HT\right)+P\left(TH\right)=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$
$P(X=2)=P\left(HH\right)=\frac{1}{4}$
Thus, the required probability distribution is as follows.
(ii) When three coins are tossed simultaneously, the sample space is $\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}$
Let $X$ represents the number of tails.
It can be seen that $X$ can take the value of $0,1,2$ or $3$.
$P(X=0)=P\left(HHH\right)=\frac{1}{8}$
$P(X=1)=P\left(HHT\right)+P\left(HTH\right)+P\left(THH\right)=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}$
$P(X=2)=P\left(HTT\right)+P\left(THT\right)+P\left(TTH\right)=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{3}{8}$
$P(X=3)=P\left(TTT\right)=\frac{1}{8}$
Thus, the probability distribution is as follows.
(iii) When a coin is tossed four times, the sample space is
$S=\{HHHH,HHHT,HHTH,HHTT,HTHT,HTHH,HTTH,HTTT,THHH,THHT,THTH,THTT,TTHH,TTHT,TTTH,TTTT\}$
Let $X$ be the random variable, which represents the number of heads.
It can be seen that $X$ can take the value of $0,1,2,3$ or $4$.
$P(X=0)=P\left(TTTT\right)=\frac{1}{16}$
$P(X=1)=P\left(TTTH\right)+P\left(TTHT\right)+P\left(THTT\right)+P\left(HTTT\right)=\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{4}{16}=\frac{1}{4}$
$P(X=2)=P\left(HHTT\right)+P\left(THHT\right)+P\left(TTHH\right)+P\left(HTTH\right)+P\left(HTHT\right)+P\left(THTH\right)=\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{6}{16}=\frac{3}{8}$
$P(X=3)=P\left(HHHT\right)+P\left(HHTH\right)+P\left(HTHH\right)+P\left(THHH\right)=\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{4}{16}=\frac{1}{4}$
$P(X=4)=P\left(HHHH\right)=\frac{1}{16}$
Thus, the probability distribution is as follows.